Timeline for Are there primes p, q such that p^4+1 = 2q^2 ?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| May 16, 2010 at 16:10 | comment | added | Franz Lemmermeyer | @Wadim: I can't see why a) cannot work. As for c), 2p^4 + 2 = (p^2+1)^2 + (p^2-1)^2 = (2q)^2. | |
| May 14, 2010 at 20:45 | comment | added | JSE | Interestingly enough, the similar equation p^2 + 1 = 2q^4 DOES have an interesting solution, namely (p,q) = (239,13). This is related to Machin's identity pi/4 = 4 arctan(1/5) - arctan(1/239). (See Ribenboim's book on Catalan Conjecture for this.) See p.7-8 of www.math.wisc.edu/~ellenber/MCAV.pdf for an "explanation" of this solution via a congruence mod 5 between a weight-2 cuspform in conductor 1024 and an Eisenstein series. | |
| May 14, 2010 at 18:28 | comment | added | Qiaochu Yuan | @Hugo: I know at least one author (Stanley) who uses P to stand for the positive integers and N to stand for the non-negative integers. | |
| May 14, 2010 at 15:42 | vote | accept | Hugo van der Sanden | ||
| May 14, 2010 at 15:26 | answer | added | user6096 | timeline score: 21 | |
| May 14, 2010 at 14:01 | answer | added | Eben Freeman | timeline score: 4 | |
| May 14, 2010 at 13:45 | comment | added | Wadim Zudilin | @Franz: This is a variation of Pell's equation, so $p^2+q\sqrt2=(1+\sqrt2)^{2k+1}$ (it can be shown by modular argument that $k$ is multiple of 4). Thus (a) can't work, (b) is used in derivation of the above formula (there are useless recursions mentioned by the author), and for (c) I have no idea of what do you mean (how to relate the equation with Pythagorean triples?) | |
| May 14, 2010 at 13:31 | comment | added | teil | Anglin's book - The Queen of Mathematics: An Introduction to Number Theory has solutions to Diophantine equations very similar to this, it might even have this one. | |
| May 14, 2010 at 13:03 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
edited title; edited title
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| May 14, 2010 at 12:32 | comment | added | Hugo van der Sanden | Yes, $\mathbb{P}$ is the primes. That is standard isn't it? | |
| May 14, 2010 at 12:31 | comment | added | Franz Lemmermeyer | If you're looking for a simple proof, you probably have to a) factor the left hand side over Z[i]; b) subtract 1 and factor the right hand side over Z[sqrt{2}], or c) multiply by 2 and use the classification of Pythagorean triples (or subtract a square and factor the right hand side, which is a difference of squares). | |
| May 14, 2010 at 12:25 | answer | added | Kevin Buzzard | timeline score: 6 | |
| May 14, 2010 at 12:13 | history | asked | Hugo van der Sanden | CC BY-SA 2.5 |